Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $t = \dfrac{7r + 21}{-3r + 3} \times \dfrac{r^2 - r}{r^2 + 10r + 21} $
Answer: First factor the quadratic. $t = \dfrac{7r + 21}{-3r + 3} \times \dfrac{r^2 - r}{(r + 3)(r + 7)} $ Then factor out any other terms. $t = \dfrac{7(r + 3)}{-3(r - 1)} \times \dfrac{r(r - 1)}{(r + 3)(r + 7)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ 7(r + 3) \times r(r - 1) } { -3(r - 1) \times (r + 3)(r + 7) } $ $t = \dfrac{ 7r(r + 3)(r - 1)}{ -3(r - 1)(r + 3)(r + 7)} $ Notice that $(r - 1)$ and $(r + 3)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ 7r\cancel{(r + 3)}(r - 1)}{ -3(r - 1)\cancel{(r + 3)}(r + 7)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $t = \dfrac{ 7r\cancel{(r + 3)}\cancel{(r - 1)}}{ -3\cancel{(r - 1)}\cancel{(r + 3)}(r + 7)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $t = \dfrac{7r}{-3(r + 7)} $ $t = \dfrac{-7r}{3(r + 7)} ; \space r \neq -3 ; \space r \neq 1 $